Problem: Let $g(x)=-8x-\cos(x)$. $g'(x)=$
Explanation: The expression for $g(x)$ includes $\cos(x)$. Remember that the derivative of $\cos(x)$ is $-\sin(x)$. Put another way, $\dfrac{d}{dx}[\cos(x)]=-\sin(x)$. $\begin{aligned} g'(x)&=\dfrac{d}{dx}[-8x-\cos(x)] \\\\ &=-8\dfrac{d}{dx}(x)-\dfrac{d}{dx}[\cos(x)] \\\\ &=-8\cdot (1)-(-\sin(x)) \\\\ &=-8+\sin(x) \end{aligned}$ In conclusion, $g'(x)=-8+\sin(x)$